01 If 11 12 21 22 we de64257ne the determinant of also denoted by det to be the scalar det 11 22 12 21 The notation 11 12 21 22 is also used for the determinant of If is a real matrix there is a geometrical interpretation of de If ID: 25600 Download Pdf

01 If 11 12 21 22 we de64257ne the determinant of also denoted by det to be the scalar det 11 22 12 21 The notation 11 12 21 22 is also used for the determinant of If is a real matrix there is a geometrical interpretation of de If

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Chapter 4 DETERMINANTS DEFINITION 4.0.1 If 11 12 21 22 , we deﬁne the determinant of , (also denoted by det ,) to be the scalar det 11 22 12 21 The notation 11 12 21 22 is also used for the determinant of If is a real matrix, there is a geometrical interpretation of de . If = ( , y ) and = ( , y ) are points in the plane, forming a triangle with the origin = (0 0), then apart from sign, is the area of the triangle OPQ . For, using polar coordinates, let cos and sin , where OP and is the angle made by the ray OP with the positive –axis. Then triangle OPQ has area OP OQ

sin , where POQ . If triangle OPQ has anti–clockwise orientation, then the ray OQ makes angle with the positive –axis. (See Figure 4.1.) Also cos and sin . Hence Area OPQ OP OQ sin OP OQ sin( OP OQ (sin cos cos sin OQ sin OP cos OQ cos OP sin 71

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72 CHAPTER 4. DETERMINANTS @ Figure 4.1: Area of triangle OPQ Similarly, if triangle OPQ has clockwise orientation, then its area equals For a general triangle , with = ( , y , i = 1 3, we can take as the origin. Then the above formula gives or according as vertices are anti–clockwise or clockwise oriented. We now give a

recursive deﬁnition of the determinant of an matrix = [ ij , n 3. DEFINITION 4.0.2 (Minor) Let ij ) (or simply ij if there is no ambiguity) denote the determinant of the ( 1) 1) submatrix of formed by deleting the –th row and –th column of . ( ij ) is called the ( i, j minor of .) Assume that the determinant function has been deﬁned for mat rices of size ( 1) 1). Then det is deﬁned by the so–called ﬁrst–row Laplace

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73 expansion det 11 11 12 12 ) + . . . + ( 1) 1+ =1 1) 1+ For example, if = [ ij ] is a 3 3 matrix, the Laplace expansion gives det 11

11 12 12 ) + 13 13 11 22 23 32 33 12 21 23 31 33 13 21 22 31 32 11 22 33 23 32 12 21 33 23 31 ) + 13 21 32 22 31 11 22 33 11 23 32 12 21 33 12 23 31 13 21 32 13 22 31 (The recursive deﬁnition also works for 2 2 determinants, if we deﬁne the determinant of a 1 1 matrix [ ] to be the scalar det 11 11 12 12 ) = 11 22 12 21 EXAMPLE 4.0.1 If is a triangle with = ( , y , i = 1 3, then the area of triangle is or according as the orientation of is anti–clockwise or clockwise. For from the deﬁnition of 3 3 determinants, we have One property of determinants that follows immediately

from the deﬁni- tion is the following: THEOREM 4.0.1 If a row of a matrix is zero, then the value of the de- terminant is zero.

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74 CHAPTER 4. DETERMINANTS (The corresponding result for columns also holds, but here a simple proof by induction is needed.) One of the simplest determinants to evaluate is that of a lowe r triangular matrix. THEOREM 4.0.2 Let = [ ij ], where ij = 0 if i < j . Then det A = 11 22 . . .a nn (4.1) An important special case is when is a diagonal matrix. If =diag ( , . . ., a ) then det . . .a . In particular, for a scalar matrix tI , we have det (

tI ) = Proof . Use induction on the size of the matrix. The result is true for = 2. Now let n > 2 and assume the result true for matrices of size 1. If is , then expanding det along row 1 gives det 11 22 . . . 32 33 . . . . . . a nn 11 22 . . .a nn by the induction hypothesis. If is upper triangular, equation 4.1 remains true and the proof is again an exercise in induction, with the slight diﬀerence that the column version of theorem 4.0.1 is needed. REMARK 4.0.1 It can be shown that the expanded form of the determi- nant of an matrix consists of ! signed products . . . a ni where ( , i

, . . ., i ) is a permutation of (1 , . . ., n ), the sign being 1 or 1, according as the number of inversions of ( , i , . . ., i ) is even or odd. An inversion occurs when > i but r < s . (The proof is not easy and is omitted.) The deﬁnition of the determinant of an matrix was given in terms of the ﬁrst–row expansion. The next theorem says that we can e xpand the determinant along any row or column. (The proof is not eas y and is omitted.)

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75 THEOREM 4.0.3 det =1 1) ij ij for = 1 , . . ., n (the so–called –th row expansion) and det =1 1) ij ij for = 1 , . . .,

n (the so–called –th column expansion). REMARK 4.0.2 The expression ( 1) obeys the chess–board pattern of signs: . . . . . . . . . The following theorems can be proved by straightforward ind uctions on the size of the matrix: THEOREM 4.0.4 A matrix and its transpose have equal determinants; that is det = det A. THEOREM 4.0.5 If two rows of a matrix are equal, the determinant is zero. Similarly for columns. THEOREM 4.0.6 If two rows of a matrix are interchanged, the determi- nant changes sign. EXAMPLE 4.0.2 If = ( , y ) and = ( , y ) are distinct points, then the line through and has equation x

y = 0

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76 CHAPTER 4. DETERMINANTS For, expanding the determinant along row 1, the equation bec omes ax by = 0 where and This represents a line, as not both and can be zero. Also this line passes through , i = 1 2. For the determinant has its ﬁrst and –th rows equal if and and is consequently zero. There is a corresponding formula in three–dimensional geom etry. If , P , P are non–collinear points in three–dimensional space, with , y , z , i = 1 3, then the equation x y z = 0 represents the plane through , P , P . For, expanding the determinant along row 1, the equation

becomes ax by cz = 0, where , b , c As we shall see in chapter 6, this represents a plane if at leas t one of a, b, c is non–zero. However, apart from sign and a factor , the determinant expressions for a, b, c give the values of the areas of projections of triangle on the ( y, z x, z ) and ( x, y ) planes, respectively. Geometrically, it is then clear that at least one of a, b, c is non–zero. It is also possible to give an algebraic proof of this fact. Finally, the plane passes through , i = 1 3 as the determinant has its ﬁrst and –th rows equal if , y , z and is consequently zero. We

now work towards proving that a matrix is non–singul ar if its determinant is non–zero. DEFINITION 4.0.3 (Cofactor) The ( i, j ) cofactor of , denoted by ij ) (or ij if there is no ambiguity) is deﬁned by ij ) = ( 1) ij

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77 REMARK 4.0.3 It is important to notice that ij ), like ij ), does not depend on ij . Use will be made of this observation presently. In terms of the cofactor notation, Theorem 4 3 takes the form THEOREM 4.0.7 det =1 ij ij for = 1 , . . ., n and det =1 ij ij for = 1 , . . ., n Another result involving cofactors is THEOREM 4.0.8 Let be an matrix. Then

=1 ij kj ) = 0 if k. Also =1 ij ik ) = 0 if k. Proof If is and , let be the matrix obtained from by replacing row by row . Then det = 0 as has two identical rows. Now expand det along row . We get 0 = det =1 kj kj =1 ij kj in view of Remark 4.0.3.

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78 CHAPTER 4. DETERMINANTS DEFINITION 4.0.4 (Adjoint) If = [ ij ] is an matrix, the ad- joint of , denoted by adj , is the transpose of the matrix of cofactors. Hence adj 11 21 12 22 nn Theorems 4.0.7 and 4.0.8 may be combined to give THEOREM 4.0.9 Let be an matrix. Then (adj ) = (det = (adj A. Proof adj ik =1 ij (adj jk =1 ij kj ik

det = ((det ik Hence (adj ) = (det . The other equation is proved similarly. COROLLARY 4.0.1 (Formula for the inverse) If det = 0, then is non–singular and det adj A. EXAMPLE 4.0.3 The matrix 1 2 3 4 5 6 8 8 9 is non–singular. For det 5 6 8 9 4 6 8 9 + 3 4 5 8 8 3 + 24 24 = 0

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79 Also 11 21 31 12 22 32 13 23 33 5 6 8 9 2 3 8 9 2 3 5 6 4 6 8 9 1 3 8 9 1 3 4 6 4 5 8 8 1 2 8 8 1 2 4 5 3 6 12 15 6 8 8 The following theorem is useful for simplifying and numeric ally evaluating a determinant. Proofs are obtained by expanding along the co rresponding row or column. THEOREM 4.0.10 The

determinant is a linear function of each row and column. For example 11 11 12 12 13 13 21 22 23 31 32 33 11 12 13 21 22 23 31 32 33 11 12 13 21 22 23 31 32 33 ta 11 ta 12 ta 13 21 22 23 31 32 33 11 12 13 21 22 23 31 32 33 COROLLARY 4.0.2 If a multiple of a row is added to another row, the value of the determinant is unchanged. Similarly for column s. Proof . We illustrate with a 3 3 example, but the proof is really quite general. 11 ta 21 12 ta 22 13 ta 23 21 22 23 31 32 33 11 12 13 21 22 23 31 32 33 ta 21 ta 22 ta 23 21 22 23 31 32 33

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80 CHAPTER 4. DETERMINANTS 11 12 13 21

22 23 31 32 33 21 22 23 21 22 23 31 32 33 11 12 13 21 22 23 31 32 33 11 12 13 21 22 23 31 32 33 To evaluate a determinant numerically, it is advisable to re duce the matrix to row–echelon form, recording any sign changes caused by ro w interchanges, together with any factors taken out of a row, as in the followi ng examples. EXAMPLE 4.0.4 Evaluate the determinant 1 2 3 4 5 6 8 8 9 Solution . Using row operations and and then expanding along the ﬁrst column, gives 1 2 3 4 5 6 8 8 9 1 2 3 15 15 1 2 15 1 2 0 1 EXAMPLE 4.0.5 Evaluate the determinant 1 1 2 1 3 1 4 5 7 6 1 2 1 1 3 4 Solution

1 1 2 1 3 1 4 5 7 6 1 2 1 1 3 4 1 1 2 1 2 2 13 0 0 1 3

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81 1 1 2 1 0 1 1 13 0 0 1 3 1 1 2 1 0 1 1 0 0 12 0 0 1 3 = 2 1 1 2 1 0 1 1 0 0 1 3 0 0 12 = 2 1 1 2 1 0 1 1 0 0 1 3 0 0 0 30 = 60 EXAMPLE 4.0.6 (Vandermonde determinant) Prove that 1 1 1 a b c = ( )( )( Solution . Subtracting column 1 from columns 2 and 3 , then expanding along row 1, gives 1 1 1 a b c 1 0 0 a b a c a c = ( )( 1 1 a c = ( )( )( REMARK 4.0.4 From theorems 4.0.6, 4.0.10 and corollary 4.0.2, we de- duce (a) det( ij ) = det (b) det( ) = det , if = 0,

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82 CHAPTER 4. DETERMINANTS (c) det( ij )

=det It follows that if is row–equivalent to , then det det , where = 0. Hence det = 0 det = 0 and det = 0 det = 0. Consequently from theorem 2.5.8 and remark 2.5.7, we have the following im portant result: THEOREM 4.0.11 Let be an matrix. Then (i) is non–singular if and only if det = 0; (ii) is singular if and only if det = 0; (iii) the homogeneous system AX = 0 has a non–trivial solution if and only if det = 0. EXAMPLE 4.0.7 Find the rational numbers for which the following homogeneous system has a non–trivial solution and solve the system for these values of + 3 = 0 ax + 3 + 2 = 0 az = 0

Solution . The coeﬃcient determinant of the system is ∆ = 2 3 3 2 6 1 2 3 0 3 + 2 0 13 18 3 + 2 13 18 = (3 + 2 )( 18) 13(2 = 2 + 6 80 = 2( + 8)( 5) So ∆ = 0 8 or = 5 and these values of are the only values for which the given homogeneous system has a non–trivial soluti on. If 8, the coeﬃcient matrix has reduced row–echelon form equal t 1 0 0 1 0 0 0

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83 and so the complete solution is z, y = 2 , with arbitrary. If = 5, the coeﬃcient matrix has reduced row–echelon form equal to 1 0 1 0 1 0 0 0 and so the complete solution is z, y , with

arbitrary. EXAMPLE 4.0.8 Find the values of for which the following system is consistent and solve the system in each case: = 1 tx (1 + + 2 = 3 Solution . Suppose that the given system has a solution ( , y ). Then the following homogeneous system = 0 tx tz = 0 (1 + + 2 + 3 = 0 will have a non–trivial solution , y , z Hence the coeﬃcient determinant ∆ is zero. However ∆ = 1 1 1 1 + 2 3 1 0 0 1 + = (1 )(2 Hence = 1 or = 2. If = 1, the given system becomes = 1 = 1 + 2 = 3 which is clearly inconsistent. If = 2, the given system becomes = 1 = 2 + 2 = 3

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84

CHAPTER 4. DETERMINANTS which has the unique solution = 1 , y = 0. To ﬁnish this section, we present an old (1750) method of solv ing a system of equations in unknowns called Cramer’s rule . The method is not used in practice. However it has a theoretical use as it re veals explicitly how the solution depends on the coeﬃcients of the augmented m atrix. THEOREM 4.0.12 (Cramer’s rule) The system of linear equations in unknowns , . . ., x 11 12 21 22 nn has a unique solution if ∆ = det [ ij = 0, namely , x , . . ., x where is the determinant of the matrix formed by replacing

the –th column of the coeﬃcient matrix by the entries , b , . . ., b Proof . Suppose the coeﬃcient determinant = 0. Then by corollary 4.0.1, exists and is given by adj and the system has the unique solution 11 21 12 22 nn 11 21 . . . 12 22 . . . . . . nn However the –th component of the last vector is the expansion of along column . Hence

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4.1. PROBLEMS 85 4.1 PROBLEMS 1. If the points = ( , y , i = 1 4 form a quadrilateral with ver- tices in anti–clockwise orientation, prove that the area of the quadri- lateral equals (This formula generalizes to a simple

polygon and is known as the Surveyor’s formula .) 2. Prove that the following identity holds by expressing the left–hand side as the sum of 8 determinants: x b y c u y v z a v b w = 2 a b c x y z u v w 3. Prove that + 1) + 2) + 1) + 2) + 3) + 2) + 3) + 4) 4. Evaluate the following determinants: (a) 246 427 327 1014 543 443 342 721 621 (b) 1 2 3 4 2 1 4 3 1 2 4 3 [Answers: (a) 29400000; (b) 900.] 5. Compute the inverse of the matrix 1 0 3 1 4 5 2 by ﬁrst computing the adjoint matrix. [Answer: 13 11 4 2 29 7 10 2 1 .]

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86 CHAPTER 4. DETERMINANTS 6. Prove that the

following identities hold: (i) b b a a b a b b 2( (ii) c b c c c a a b a a = 2 7. Let = ( , y , i = 1 3. If , x , x are distinct, prove that there is precisely one curve of the form ax bx passing through , P and 8. Let 1 1 2 3 Find the values of for which det = 0 and hence, or otherwise, determine the value of for which the following system has more than one solution: = 1 + 3 kz = 3 ky + 3 = 2 Solve the system for this value of and determine the solution for which has least value. [Answer: = 2; = 10 21 , y = 13 21 , z = 2 21.] 9. By considering the coeﬃcient determinant, ﬁnd all

ration al numbers and b for which the following system has (i) no solutions, (ii ) exactly one solution, (iii) inﬁnitely many solutions: bz = 3 ax + 2 = 2 + 2 = 1 Solve the system in case (iii). [Answer: (i) ab = 12 and = 3, no solution; ab = 12, unique solution; = 3 , b = 4, inﬁnitely many solutions; , y , with arbitrary.]

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4.1. PROBLEMS 87 10. Express the determinant of the matrix 1 1 2 1 1 2 3 4 2 4 7 2 + 6 2 2 6 t t as as polynomial in and hence determine the rational values of for which exists. [Answer: det = ( 2)(2 1); = 2 and .] 11. If is a 3 3 matrix over

a ﬁeld and det = 0, prove that (i) det (adj ) = (det (ii) (adj det = adj( 12. Suppose that is a real 3 3 matrix such that (i) Prove that ) = (ii) Prove that det 1. (iii) Use (i) to prove that if det = 1, then det( ) = 0. 13. If is a square matrix such that one column is a linear combinatio n of the remaining columns, prove that det = 0. Prove that the converse also holds. 14. Use Cramer’s rule to solve the system + 3 = 1 + 2 = 4 3. [Answer: = 2 , y = 3 , z = 4.] 15. Use remark 4.0.4 to deduce that det ij det ) = t, det ij ) = 1 and use theorem 2.5.8 and induction, to prove that det( BA

) = det det A, if is non–singular. Also prove that the formula holds when is singular.

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88 CHAPTER 4. DETERMINANTS 16. Prove that c a b a a b a c a a a a a c a a a a b a (2 )(4 +2 17. Prove that 1 + 1 + 1 + 1 + = 1 + 18. Let ). If , prove that det = 0 if is odd and 1 + 1 = 0 in 19. Prove that 1 1 1 1 1 1 1 r r 1 1 r r r = (1 20. Express the determinant bc a ca b ab c as the product of one quadratic and four linear factors. [Answer: ( )( )( )( )( bc ac ab ).]

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